Calculate the standard cell emf for galvanic cells with the cell reaction

2H2(g) + O2(g)→2H2O(ℓ)in acidic solution.
Answer in units of V

1.23 V

To calculate the standard cell emf for a galvanic cell, we need to use the standard reduction potentials for the half-reactions involved in the cell reaction. The standard reduction potential values are typically given in tables.

The cell reaction given is:
2H2(g) + O2(g) → 2H2O(ℓ) (in acidic solution)

Now, let's break down the cell reaction into two half-reactions:
1. Oxidation half-reaction: H2(g) → 2H+(aq) + 2e-
2. Reduction half-reaction: O2(g) + 4H+(aq) + 4e- → 2H2O(ℓ)

To calculate the standard cell emf, we need to find the standard reduction potentials (E°) for these half-reactions. Let's assume that the standard reduction potential for the hydrogen half-reaction is E°(H2) and the standard reduction potential for the oxygen half-reaction is E°(O2).

Then, the standard cell emf (E°cell) can be calculated using the formula:
E°cell = E°(reduction) - E°(oxidation)

From standard reduction potential tables, we can find that:
E°(H2) = 0 V (The reduction potential of H2 is defined as 0 V)
E°(O2) = +1.23 V

Now, let's calculate the standard cell emf:
E°cell = E°(reduction) - E°(oxidation)
E°cell = E°(O2) + E°(H+) - E°(H2)
E°cell = +1.23 V + 0 V - 0 V
E°cell = +1.23 V

Therefore, the standard cell emf for the given galvanic cell reaction is +1.23 V.

Look up these potentials.

O2 + 4H^+ + 4e ==> 2H2O
2H2 ==> 4H^+ + 4e
----------------------
Add Eo1(as written) + Eo2(as written)

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